Question: Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{3x^2 + 39x + 90}{-9x^3 + 900x}$
First factor out the greatest common factors in the numerator and in the denominator. $ a = \dfrac {3(x^2 + 13x + 30)} {-9x(x^2 - 100)} $ $ a = -\dfrac{3}{9x} \cdot \dfrac{x^2 + 13x + 30}{x^2 - 100} $ Simplify: $ a = - \dfrac{1}{3x} \cdot \dfrac{x^2 + 13x + 30}{x^2 - 100}$ Next factor the numerator and denominator. $ a = - \dfrac{1}{3x} \cdot \dfrac{(x + 10)(x + 3)}{(x + 10)(x - 10)}$ Assuming $x \neq -10$ , we can cancel the $x + 10$ $ a = - \dfrac{1}{3x} \cdot \dfrac{x + 3}{x - 10}$ Therefore: $ a = \dfrac{ -x - 3 }{ 3x(x - 10)}$, $x \neq -10$